a flock of seagulls has decided to mount an organized response to A flock of sea

Question

a flock of seagulls has decided to mount an organized response to A flock of seagulls has decided to mount an organized response to the human overpopulation of their favorite beach. One tactic popular among the innovative radicals is bombing the sunbathers with dams. A gull dives with a speed of 16 m/s, at an angle of 40 degree below the horizontal. He releases a projectile when his vertical distance above his target, a sunbather's bronzed tummy, is 8.5 m, and scores a bull's-eye. (a) Where is the sunbather in relation to the gull at the instant of release? (W How long is the projectile in the air? (c) What is the velocity of the projectile upon impact?

Explanation / Answer

PROJECTILE

1)

along horizontal

initial velocity vx = v*cos(theta)

ax = 0

from equation of motion

x = vx*T+ 0.5*ax*T^2

x = v*costheta*T

T = x/(v*cos(theta))......(1)

along vertical

vy = -v*sin(theta)

acceleration ay = -g = -9.8 m/s^2

from equation of motion

y-y0 = -vy*T + 0.5*ay*T^2

y-y0 = -(v*sin(theta)*x)/(v*cos(theta)) - (0.5*g*x^2)/(v^2*(cos(theta))^2)

y-y0 = -x*tan(theta) - ((0.5*g*x^2)/(v^2*(cos(theta))^2))

yo = 8.5 m

y = 0

theta = 40

v = 16 m/s

x = ?

(a)

-8.5 = -xtan40 - (0.5*9.8*x^2/(16^2*(cos40)^2)

x = 7.78 m <<<<---------answer

(b)

T = x/(v*costheta) = 7.78/(16*cos40) =0.635 s

(c)

vx = v*cotheta = 16*cos40 =12.3 m/s

vy = voy - g*T = -(16*sin40)-(9.8*0.635) = -16.5 m/s

speed v = sqrt(vx^2+vy^2)

v = sqrt(12.3^2+16.5^2) = 20.6 m/s <<<-----answer

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